At the same time contract diagonalization
Theorem:
set is $ A $ $ $ n-order real symmetric positive definite matrices, $ B $ is a real symmetric matrix of the same order, certainly exist invertible matrix $ C $, such that
\ [C'AC = I_n, C'BC = diag \ {\ lambda_1, \ cdots, \ lambda_n \} \]
where $ \ {\ lambda_1, \ cdots , \ lambda_n \} $ matrix $ a ^ {- 1} B $ eigenvalues.
Proof:
Since $ a $ is positive definite, the reversible matrix $ P $ is present, so that $ P'AP = I_n $. $ P'BP $ Since the matrix is a real symmetric matrix, $ Q $ orthogonal matrix is present, so that the $ Q '(P'BP ) Q = diag \ {\ lambda_1 , \ cdots, \ lambda_n \} $.
makes $ C = PQ $, is satisfied $ C'BC = diag \ {\ lambda_1 , \ cdots, \ lambda_n \} $. Because of
\ [ C '(\ lambda AB) C = \ lambda I_n-C'BC = diag \ {\ lambda- \ lambda_1, \ cdots, \ lambda- \ lambda_n \} \]
is $ \ lambda_i $ polynomial $ | \ lambda AB | $ root, and $ a $ reversible, it is also to $ | \ lambda I_n-a ^ {- 1} B | $ the root.
demonstrate the use of a few examples of the above-described Theorem is:
example. 1:
set to $ n $ a $ real positive definite symmetric matrix order $, $ B $ is the same order Semidefinite real symmetric matrices, the $ | a + B | \ geq | a | + | B | $, necessary and sufficient conditions are established equals $ B = O $ .
Proof:
by Theorem known, the presence of reversible matrix $ C $, such that
\ [C'AC = I_n, C'BC = diag \ {\ lambda_1, \ cdots, \ lambda_n \} \]
Since the semi-definite $ B $, $ C'BC $ it is also semi-definite, so $ \ lambda_i \ geq 0 $ due.
\ [| C '|| A || B + C | = | C'AC C'BC + | = (1+ \ lambda_1) \ cdots (1+ \ lambda_n)
\ GEQ 1+ \ lambda_1 \ cdots \ lambda_n = | C'AC | + | C'BC | = | C '| (| A | + | B |) | C | \]
therefore, $ | A + B | \ geq | A | + . | B | $ equality if and only if all of the $ \ lambda_i = 0 $, if and only if O = $ B $.
Remark, the:
the above example can be generalized as: $ a $ is a positive semi-definite matrix only. using the perturbation method can be proven.
Example 2:
set $ A, D $ is a square matrix, $ M = \ left (\ the begin Array} {} {CC
A \\ B &
B '& D \ Array End {} \ right) $ real symmetric positive definite matrix Prove:
$ | M | \ Leq | A || D | $, and equals if and only if $ B = O $ was founded upon.
Proof:
Since $ A $ reversible, carried out on $ M $ contract early transformation:
\ [\ left (\ the begin Array} {} {CC
A \\ B &
B '& D \ Array End {} \ right) \ rightarrow
\ left (\ the begin Array} {} {CC
A & B \\
O & D ^ {-B'A -. 1} B \ Array End {} \ right) \ rightarrow
\ left (\ the begin Array} {} {CC
A & O \\
O & D-B'A ^ {-}. 1 B \ Array} {End \ right) \]
. therefore, $ D-B'A ^ {- 1 } B $ clearly positive definite matrix
\ [\ left | \ the begin Array} {} {CC
A \\ B &
B '& D \ end {array} \ right | = | A || D-B'A ^ {- 1} B | \]
because $ D = (D-B'A ^ {- 1} B) + B'A ^ {-1} B $, and $ B'A ^ {- 1} B $ semi positive definite matrices.According to the above Example 1 art,
\ [| D | \ geq | D-B'A ^ {- 1} B | + | B'A ^ {- 1} B | \ geq | D-B'A ^ {- 1} B | \]
, etc. No. if and only if $ B'A ^ {- 1}. B = O $ isobutyl provided there is a non real matrix $ C $, such that $ A ^ {- 1} = C'C $, the $ O = B ' A ^ {- 1}.. B = (CB) '(CB) $, traces have taken $ CB = O $ $ C $ nonsingular since, so $ B = O $ such that equality holds if and only if $ B . = O $ then the conclusion is proved.
Example. 3:
set $ A, B $ $ are $ n-order real positive definite symmetric matrix, confirmation
\ [| A + B | \ geq 2 ^ n | A | ^ {\ dfrac { 1} {2}} | B | ^ {\ dfrac {1} {2}} \]
Sufficient conditions equality holds is B = $ a $.
Proof:
by Theorem known, the presence of reversible matrix $ C $, such that
\ [C'AC = I_n, C'BC = diag \ {\ lambda_1, \ cdots, \ lambda_n \} \]
Because $ B $ positive definite, the definite $ C'BC $ to $ \ lambda_i> 0 $. Since
\ [| C '|| A || B + C | = | C'AC C'BC + | = (1+ \ lambda_1) \ cdots (1+ \ lambda_n)
\ 2 ^ n-GEQ \ sqrt {\ lambda_1 \ cdots \ lambda_n} ^ n-2 = | C'AC | ^ {\ dfrac. 1} {2} {} | C'BC | ^ {\ dfrac. 1} {2} {}
= | (n-2 ^ | C '| A | ^ {\ dfrac {1 } {2}} | B | ^ {\ dfrac {1} {2}}) | C | \]
Therefore, $ | A + B | \ geq 2 ^ n | A | ^ {\ dfrac {1} {2}} | B |. ^ {\ Dfrac {1} {2}} $ equality if and only if All $ \ lambda_i = 1 $, if and only if a = B $ $.
Remark, the:
the above example can be generalized to:. $ a, B $ semi-definite matrix using only perturbation method . prove to
Example 4 :
If $ A, B $ are positive definite matrices, and $ $ AB semi positive definite matrix, confirmation:. $ B ^ {- 1 } -A ^ {- 1} $ semi-definite matrices
Proof:
by Theorem known, the presence of $ C $ invertible matrix, such that
\ [C'AC = I_n, C'BC = diag \ {\ lambda_1, \ cdots, \ lambda_n \} \]
because $ B $ positive definite, the definite $ C'BC $ to $ \ lambda_i> 0 $. Because of
\ [C '(AB) C = diag \ {1- \ lambda_1, \ cdots, 1- \ lambda_n \} \]
and $ $ Semidefinite AB, the $ \ lambda_I \ leq 1 $ to $ \ lambda_i ^ {- 1} \ geq 1 $ due.
\ [C ^ {-. 1} A ^ {-. 1} (C ') ^ {-. 1} = I_n, C ^ {-. 1} B ^ {-1} (C ') ^ {- 1} = diag \ {\ lambda_1 ^ {- 1}, \ cdots, \ lambda_n ^ {- 1} \} \]
is
\ [C ^ {- 1} (B ^ {- 1} -A ^ { - 1}) (C ^ {- 1}) '= diag \ {\ lambda_1 ^ {- 1} -1, \ cdots, \ lambda_n ^ {- 1} -1 \} \]
Obviously semi-definite matrix, then $ B ^ {- 1} -A ^ {- 1} $ semi positive definite matrices.
Example. 5:
set $ A, B $ a $ $ n-order real symmetric matrices, where definite $ A $ $ and $ B $ and $ AB are semi positive definite matrices confirmation:. $ | \ lambda AB | = 0 $ $ all fell on the root [0,1] $, and the $ | a | \ geq | B | $.
Proof:
by Theorem known, the presence of reversible matrix $ C $, such that
\ [C'AC = I_n, C'BC = diag \ {\ lambda_1, \ cdots, \ lambda_n \} \]
where $ \ lambda_i $ matrix $ a ^ {- 1} B $ feature value, namely $ | \ lambda AB | $ $ B $ root due Semidefinite, the $ C'BC $ Semidefinite to $ \ lambda_i \ geq 0 $. . because $ $ Semidefinite AB, the $ C '(AB) C = diag \ {1- \ lambda_1, \ cdots, 1- \ lambda_n \} $ Semidefinite, i.e. $ \ lambda_1 \ leq 1 $ so $ |. \ lambda AB | $ root of all $ \ lambda_i $ left all $ [0,1] and the $ $ |. a ^ {- 1 } B | = \ lambda_1 \ cdots \ lambda_n \ leq 1 $, so $ | a | \ geq | B | $.
Theorem:
set $ A, B $ $ n-order of $ Semidefinite real symmetric matrices, $ C $ invertible real matrix is present, such that
\ [C'AC = diag \ {1 , \ cdots, 1,0 \ cdots, 0 \}, C'BC = diag \ {\ lambda_1, \ cdots, \ lambda_r, \ R & lt lambda_ {+}. 1, \ cdots, \ lambda_n \} \]
Proof:
Since $ A $ semi-definite matrix, the presence of reversible matrix $ P $, such P'AP = $ \ left (\ the begin Array} {} {CC
I_r & O \\
O & O \ Array End {} \ right) $, and $ P'BP = \ left (\ {} {CC Array the begin}
B_ {}. 11 & 12 is B_ {} \\
B_ 21 is {{} & B_ 22 is} \} End {Array \ right) $ semi positive definite matrices. by quadratic theory known, $ r (B_ {21} B_ {22 }) = r (B_ {22}) $. That $ B_ {21} $ all can be expressed as a column vector $ B_ {22} $ linear combination of column vectors, there is a real matrix $ M $, so $ B_ {21} = B_ { 22} M $ matrix $ P'AP, P'BP $ a contract conversion:.
\ [\ left (\ the begin Array} {} {CC
I_r & - M '\\
O & {NR} of I_ \ Array End {} \ right)
\ left (\ the begin Array} {} {CC
B_ {}. 11 & 12 is B_ {\\}
B_ {} & B_ 21 is 22 is} {\ end {array} \ right)
\left(\begin{array}{cc}
I_r & O\\
-M' & I_{n-r}\end{array}\right)=
\left(\begin{array}{cc}
B_{11}-M'B_{22}M & O\\
O & B_{22}\end{array}\right)\]
\[\left(\begin{array}{cc}
I_r & -M'\\
O & I_{n-r}\end{array}\right)
\left(\begin{array}{cc}
I_r & O\\
O & O\end{array}\right)
\left(\begin{array}{cc}
I_r & O\\
-M' & I_{n-r}\end{array}\right)=
\left(\begin{array}{cc}
I_r & O\\
O & O\end{array}\right)\]
由于$B_{11}-M'B_{22}M$与$B_{22}$均为半正定矩阵，则存在正交矩阵$Q_1,Q_2$，使得
\[Q_1'(B_{11}-M'B_{22}M)Q_1=diag\{\lambda_1,\cdots,\lambda_r\},
Q_2'B_{22}Q_2=diag\{\lambda_{r+1},\cdots,\lambda_n\}\]
So P = C $ \ left (\ the begin Array} {} {CC
I_r \\ O &
-M 'of I_ & {NR} \} End {Array \ right)
\ left (\ the begin Array} {} {CC
Q_1 & \\ O
O & Q_2 \ Array End {} \ right) $, $ C $ the real reversible matrix, such that
\ [C'AC = diag \ {1 , \ cdots, 1,0 \ cdots, 0 \}, = diag C'BC \ {\ lambda_1, \ cdots, \ lambda_r, \ R & lt lambda_ {+}. 1, \ cdots, \ lambda_n \} \]
Corollary:
set $ A, B $ $ are $ n-order real Semidefinite symmetric matrix, then
(1) $ a + B $ and sufficient conditions for positive definite matrices is the presence of $ n-$ linearly independent column vectors $ \ overrightarrow {\ alpha} _1 , \ cdots, \ overrightarrow {\ alpha} _n $ and the index set $ I \ subseteq \ {1,2, \ cdots, n \} $, such that
\ [\ overrightarrow {\ alpha} _i'A \ overrightarrow {\ alpha} _j = \ overrightarrow {\ alpha} _i ' B \ overrightarrow {\ alpha} _j = 0 (\ forall i \ neq j), \ overrightarrow {\ alpha} _i'A \ overrightarrow {\ alpha} _i> 0 (\ forall i \ in I), \ overrightarrow {\ alpha} _j'B \ overrightarrow {\ alpha } _j>0(\forall j\in I)\]
(2) $ R & lt (A, B) = R & lt (A + B) $
(. 3) $ A + B $ is the necessary and sufficient conditions for positive definite matrices is $ R & lt (A, B) = n-$
Proof:
proved by the Theorem It is obvious.
Reference: Advanced Algebra - Study Guide Book (Third Edition) (Xieqi Hong)